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Lecture 4Applications of diffusion4.1 Cooling of the earthLast lecture, we discussed the error function as a functional solution to thediffusion equation and discussed its general properties that made it useful,namely that erf(0) = 0 and that limx→0erf(x) = 1. Remember that the way thatwe are usually going to “solve” the diffusion equation is simply by trying tofind one of the known functional forms (or sum of functions, since we knowthat sums of solutions are solutions). So anytime we have initial conditionsthat are matched by some sort of constant along a boundry plane, we cantry to match it with an error function. As the best example of this type ofsolution, let’s consider the cooling of the earth in the absence of any internalheat sources.We can make a simplistic model of the earth by assuming it is initiallyat a temperature of 2000 K (the initial heat is from the initial gravitationalcollapse: potential energy has been converted to heat energy), but that thesurface is in equilibrium with the solar radiation and stays at a temperature of300 K, so the earth cools. What is the temperature with depth as a functionof time? We simply want a functional solution to the diffusion equation whereone boundry stays fixed and everything else is allowed to wander. Considerthen the solutionT (z, t) = Ts+ T∞erf(z/2√kt)where Tsis the 300 K surface temperature and T∞is the initial tempoera-ture everywhere else (actually it is the initial temperature minus the surfacetemperature).37Figure 4.1: The temperature of the earth as a a function of depth and time,assuming no internal sources and a fully granite earth.Let’s make sure this fits everything we have as initial conditions. Ast = 0, the value of the error function goes to 1 for every z except for z = 0,where the value goes to 0. Thus this function fits the initial condition, andwe know it’s a solution of the diffusion equation, so it must be the answer.Assuming that the whole of the earth is granite, with k = 10−6m2/s, wecan plot the solution, as seen in Figure 4.1. The whole earth graduallycools to the temperature of the surface.Lord Kelvin was the first to try to use this solution to come up with the(incorrect) age of the earth. How did he do it? The most mathematicallyobvious way would be to drill several thousand kilometers down and try todetermine where the temperature plateaus. This doens’t work in practice,though, but from the earliest days of mining it was known that a “geothermalgradient” of about 10 to 50 C per kilometer existed. Figure 4.1 shows thegradient found in our solution to the diffusion equation (in practice, whatis really plotted is the difference in temperature at the surface and at 1km depth). You can see that the earth’s thermal gradient is reached after38Figure 4.2: The thermal gradient in the first kilometer for the cooling earthas a function of time.something like 107years, thus that is the age of the earth, says Lord Kelvin!(Lord Kelvin really said 94 × 106years to get down to 10 C/km, whichis remarkably close to what we calculated). What did he do wrong? Thestandard answer is that, of course, the earth is heated from within by theradioactive decay of elements. But other things that change the answer alotthat he didn’t consider are chemical changes, phase changes, and the increasein melting point with pressure. (The problem was solved corrected in 1940by Jeffreys who got a cooling time of about 2 billion years).4.1.1 Separating the functionA general class of solutions that will lead to an interesting particular solutioncan be had by a nice general method called separating the function. Sincewe are merely looking now for any solution to the diffusion equation, whynot consider one of the formT (x, t) = g(x)f(t) + c,39where g and f are completely independent functions and c is just a constant(which we can easily see from the equation will always be able to be addedto any solution). That is, the spatial and temporal part of the solution arecompletely independent. Clearly, none of the solutions we have examined sofar fit the bill. But we can generally solve for g and f in the following way.Because T (x, y) is a solution to the diffusion equation,ρcg∂f∂t= kf∂2g∂x2or, separating the functions completely,∂f∂tf=kρc∂2g∂x2g.Now, this equation must be true for any value of x and for any value of t.The only way for this condition to hold is if both sides of the equation areequal to the same constant, which we will call ω. We then get two equations:∂f∂t= ωf and (4.1)∂2g∂x2=ρckωg. (4.2)We now solve these two equations separately. The first is a simple firstorder linear differential equation with solutionf = f0exp(ωt).The second is a second order differential equation, but we can see byinspection that the simple solution to the equation isg = g0exp(rωρckx),where g0is the constant of integration.The full solution is the product of f and g (plus the constant), soT (x, t) = c1exp(ωt +rωρckx) + c2, (4.3)where we have combined the constants f0and g0into simply c1= f0g0.40Figure 4.3: A plot of equation 4.3 as a function of time with ω = 1 andρck= 1.Now that we have found a solution, let’s find the problem. We will takethe convenient step of setting c2= 0. What is the steady state solution to thisproblem? This is a trick question, of course, because there is no steady statesolution. The positive exponential in t ensures that whatever the solutionis, it grow steadily in time. SO what is the solution? Figure 4.1.1 showsthe function for a variety of times. Can you tell what (unphysical) situationthat this function is the solution for? Think carefully about heat flow in thisexample, including where it comes from and where it is going.4.1.2 Imaginary solutionsEquation 4.3 was perhaps not a very useful solution, as nothing in real lifeincreases in temperature exponentially forever. It is possible, however, todecrease in temperature exponentially forever, and if we had simply assumedthat ω was negative, we would have had nice exponential decay. Let’s writethis solution out with the numbers explicitly negative:41T (x, t) = c1exp(−ωt +s−ωρckx) + c2. (4.4)In the spatial part of the solution, we now have a square root of a negativenumber, so we will have an imaginary part, which we can rewrite asT (x, t) = c1exp(−ωt + irωρckx) + c2. (4.5)You have all seen this form of imaginary exponential before and knowthatexp(ikx) = cos(ikx) + i sin(ikx).Have you ever thought


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CALTECH GE 108 - Applications of diffusion

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